Question

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and increases into quadrant 1. It goes through (negative 1, 0.25), (0, 0.5), (1, 1).
Which exponential function is represented by the graph?

f(x) = 2(one-half) Superscript x
f(x) = One-half(2)x
f(x) = One-half (one-half) superscript x
f(x) = 2(2)x

Answer 1

Answer:

           $\large\boxed{f(x)=(1/2)2^x}$

Explanation:

You can just verify the outputs of the four functions for x = -1, x = 0.25, and x = 1.

1. $f(x)=2(1/2)^x$

$x=-1\implies f(x)=2(1/2)^{-1}=2(2)=4$

Then, this function does not go through (-1, 0.25)

2. $f(x)=(1/2)2^x$

$x=-1\implies f(x)=(1/2)(2)^{-1}=(1/2)(1/2)=1/4=0.25\\ \\ x=0\implies f(x)=(1/2)(2)^0=1/2=0.5\\ \\ x=1\implies f(x)=(1/2)(2)^1=(1/2)(2)=1$

Hence, this function goes through the 3 points.

Also, you can verify that it approches y = 0 in quadrant 2, because when x approaches a very large negative number ( - ∞), $2^x$ becomes very small ( approaches zero). Therefore, this function meets all the requirements: it approaches y = 0 in quadrant 2, increases into quadrant 2, and goes through the three given points.

Answer 2

Answer:

its B

Step-by-step explanation: